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GILBERT STRANG: OK, what I want
to do today is show you
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two different ways that
derivatives are used.
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In one of them, the problem is
to find a close approximation
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to the value f at a point x.
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f of x.
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The second application
is to solve an
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equation, where often--
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and I use a different letter,
capital F, just because it's a
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different function and there
will be different
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examples for this one.
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So I just chose capital F
to keep them separate.
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This is the problem of
solving an equation.
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And Newton had an idea and
it's survived all these
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centuries and it's still
the good way.
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OK, what's this based
on, both of them?
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That's the point.
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They're both based
on the same idea.
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Suppose at a point, which is
near the x we want, or near
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the solution to this problem, at
some point, let me call it
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a, suppose we know the slope,
the derivative, at that point.
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So I'm using f prime
for the derivative.
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At that point, well, we know
what the definition is.
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But I'm also supposing that
we've got that number.
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And now I want to use this
knowledge of the slope at that
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nearby point a to come close to
the solution, the f of x,
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or the x there.
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OK, well you remember this is
delta f divided by delta x.
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You recognize that before we
take the limit this is two
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nearby points, delta x apart.
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Their values, the values
of f at those points.
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It's the change in f divided
by the change in x--
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that's what the derivative
is--
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as one point approaches
the point a
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where we know the slope.
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Here's the idea.
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I'm just going to erase
that stuff.
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Well, now I won't have an
equal sign anymore.
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I'll have an approximately
equal.
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So the slope--
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this is delta f over delta x--
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is not the same as df
dx, which is this.
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But if x is close to a this will
normally be close to the
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correct slope, the instant
slope at the point.
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I'm just going to use this
approximation to find a good
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approximation for f of x.
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So in this application on the
left I know the x and I want
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to find what is f
at that point?
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So I look at this.
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I multiply up by x minus a.
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I move the f of a to
the other side.
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And what do I have?
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I still have an approximation
sign.
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So when I move the f of a to the
other side, there it is.
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And then, the other part is x
minus a times f prime at a.
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That's my formula.
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Let me just talk about that
formula for a minute and then
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give examples.
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What that says is that if I want
to find the value of f at
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a nearby point, a good thing
to do is use this linear
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approximation.
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I use the word "linear"
because the graph
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of that is a line.
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I'm following a straight line
instead of following the
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curved graph.
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That's the message of
today's lecture.
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Follow the line.
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So it's a line.
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It starts out at the correct
point at x equal a.
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It has the corrects slope
f prime of a.
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And if I don't go too far
that line won't be too
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far from the curve.
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Good.
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You'll see it now in examples.
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Let me get the corresponding
idea here.
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Now I have to remember to use
capital F. So let me create
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the formula that helps to solve,
approximately solves,
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this equation.
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What's the difference?
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Now x is what I'm looking for.
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x is what I'm looking for and
F of x is what I know.
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I know it's to be 0.
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So I'm going to use this with
capital F. I know F of x
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should be 0 and x
is the unknown.
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So can I just move that equation
around again to get
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an equation?
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I'll bring the x minus
a up as I did before.
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And I have to use an
approximation sign as always.
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The F of x is 0, so I have minus
F of a, and then I'm
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dividing by F prime of a.
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There you have it.
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That is Newton's insight.
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Newton and then somebody named
Raphson helped out, made it
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work for general functions.
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And once again, I forgot.
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I should be using
capital F there.
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Because in the right side of
the board I'm calling the
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function capital F. This is the
little movement of x away
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from a, which will bring us
closer to the answer.
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You'll see is this formula
work in a picture.
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So I'm ready for an example
starting with approximation.
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So here's my problem.
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Here's my example.
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Find the square root of--
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so my problem is going to be
find the square root or
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approximate square root of 9.
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I'm going to shift away
from 9 a little bit.
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9.06.
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OK, how does that fit
this example?
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My function is the square root
function, x to the 1/2.
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It's derivative, f prime.
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I know the derivative of
that is 1/2 x to one
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lower power minus 1/2.
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So that's 1 over 2
square root of x.
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Good.
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I know my function.
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I know its derivative.
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Now I pick a point a, which
is close and easy.
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So close, I'm looking for a
point a, which is near 9.06
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and has a nice square root.
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Well, 9.
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So I'll choose a to be 9.
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The correct value f at
the point a is the
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square root of 9.
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3.
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That's the point.
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We can evaluate the function
easily at 9.
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And f prime at a is 1 over
2 square root of a.
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9.
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Square root of 9.
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So what do I have for that?
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That's the easy number, 3.
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That's 1/6.
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In other words, I know what's
happening at x equal 9.
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And that's a particular point
I'm calling a and I'm working
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from there.
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Then what does my approximation
say?
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It says that f at this
nearby point.
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So x is 9.06.
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This is the x where I want
to know the square root.
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So this is saying that the
square root of 9.06--
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that's my f of x--
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is approximately f at a.
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That's the square root of 9.
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That's the 3.
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So far reasonable.
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That's the approximation I
started with just using the 9.
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But now I'm improving
it by the difference
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between x minus 3.
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Oh, what is x?
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So I'm plugging in x is 9.06.
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That's where I really want
the square root.
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That's my x.
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Minus a, which is the 9.
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Times f prime, which we
figured out as 1/6.
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That's the linear approximation
following this
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line to this number,
which is what?
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3 plus--
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That's 0.06 divided by 6.
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That's 3.
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What do I have there?
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This difference is 0.06.
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Divided by 6 and its 0.01.
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That's the approximation.
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Closer than 3.
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That comes from following
the line.
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Let me draw a graph to show
you what I mean by
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following the line.
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Here is my square
root function.
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The square root looks
something like that.
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And here is the point x equals
9, where I know that the
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height is 3.
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What else do I know?
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Here is 9.06, a little
further over, and I'm
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looking for that point.
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I'm looking for the square
root of 9.06.
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So this was 9 here and
this was 9.06.
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And how am I getting close
to that point?
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Well, I'm not going to follow
the curve to do square root of
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9.06 exactly, any
more than your
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calculator or computer does.
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I'm going to follow this line.
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So that line is the
tangent line.
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It's the line that goes through
the right thing at a
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with the right slope.
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So you can see that by
following the line
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that gave me the 3.
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And then here is the little--
you see what I'm--
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I'm missing by a very small
amount, practically
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too small to see.
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I'm picking the point on the
line and that was this across,
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this delta x was the 0.06.
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The little tiny bit
here is the 0.06.
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And this delta f is the
little piece that I
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added on the 0.01.
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How did that come from?
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It came from the fact that
the correct slope is 1/6.
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If I go over 0.06, I
should go up 0.01.
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So you see what I'm doing?
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I'm taking that point as my
close approximation to the
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square root.
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Closer than 3 to the square
root of 9.06.
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OK.
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That's a first example.
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I'll give a second one.
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But first, I'd like to give an
example that's like this one
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of Newton's method.
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May I change now to
Newton's method?
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So I want to create
an equation.
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And actually, I want it to
solve the same problem.
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So I'm going to take my function
to be x squared minus
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9.06 and I'll set that to 0.
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I'm just keeping
my two examples
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close because the answer--
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in both problems, I'm looking
for the square root of 9.06.
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Of course, that's the solution
to this equation, the square
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root of 9.06.
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OK again, I pick a point a
close to the solution.
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And again, I'm going
to take a to be 3.
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So 3 is close to the
correct solution.
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The correct x is the square
root of 9.06.
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But I'm starting close to it.
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OK, at that point I
figure out F of a.
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Newton wants to know
F of a, the value.
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And Newton also wants
to know the slope.
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So the value at a is--
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3 squared is 9.
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9 minus 9.06 is minus 0.06.
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F prime of a is-- what's
the derivative of F?
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Capital F. Well, the derivative
is certainly 2x.
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And at the point a it's 6.
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This is the 2x.
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2a.
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This is the 2a because I'm
evaluating the slope
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at the point a.
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Actually, if you want a
picture, it's quite
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interesting to see
the picture.
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Let me graph this f of x now.
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What does my function
f of x look like?
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x squared.
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That's a parabola going up.
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It starts below here.
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It starts somewhere down here
and it curves up like so.
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So this is the point I want.
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There is the square
root of 9.06.
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00:15:21,370 --> 00:15:26,060
I've graphed a function
x squared minus 9.06.
261
00:15:26,060 --> 00:15:29,140
That's the point I'd
like to find.
262
00:15:29,140 --> 00:15:32,740
I'm not expecting to
find it exactly.
263
00:15:32,740 --> 00:15:37,200
What I do is I have
a nearby point 3.
264
00:15:37,200 --> 00:15:40,490
265
00:15:40,490 --> 00:15:47,130
At 3 I do know the exact value
and I know the slope.
266
00:15:47,130 --> 00:15:52,080
267
00:15:52,080 --> 00:15:52,175
Ha.
268
00:15:52,175 --> 00:15:57,750
Even my bad art is telling me
what's going to happen.
269
00:15:57,750 --> 00:15:59,980
Let me write that 3 better.
270
00:15:59,980 --> 00:16:02,490
a equals 3.
271
00:16:02,490 --> 00:16:04,770
This is where I know
what's happening.
272
00:16:04,770 --> 00:16:11,320
I know the value of F. It's
actually pretty small, but
273
00:16:11,320 --> 00:16:12,780
it's negative.
274
00:16:12,780 --> 00:16:14,510
I know the value of the slope.
275
00:16:14,510 --> 00:16:20,010
So this is like I've blown
up this picture here.
276
00:16:20,010 --> 00:16:23,060
277
00:16:23,060 --> 00:16:31,640
This F is the negative 0.06.
278
00:16:31,640 --> 00:16:37,550
Is the value of F.
The slope is 6.
279
00:16:37,550 --> 00:16:41,060
And what's the x?
280
00:16:41,060 --> 00:16:46,160
What is my improved guess
at the solution?
281
00:16:46,160 --> 00:16:49,610
My improved guess, I don't
follow that curve because that
282
00:16:49,610 --> 00:16:51,120
would be perfect.
283
00:16:51,120 --> 00:16:53,240
But curves are hard to follow.
284
00:16:53,240 --> 00:17:01,140
I'll follow the straight line
and that's my next x.
285
00:17:01,140 --> 00:17:03,460
My better x.
286
00:17:03,460 --> 00:17:07,170
You see, that's a lot better.
287
00:17:07,170 --> 00:17:09,089
Let me do the numbers
here and you'll see
288
00:17:09,089 --> 00:17:10,329
that it's a lot better.
289
00:17:10,329 --> 00:17:12,930
What is the x that Newton's
method gives us?
290
00:17:12,930 --> 00:17:16,109
I'm using Newton's
formula here.
291
00:17:16,109 --> 00:17:22,710
So Newton's formula says that
x minus the a is 3.
292
00:17:22,710 --> 00:17:27,589
And Newton's formula is going
to take equal to get an
293
00:17:27,589 --> 00:17:29,440
approximate x.
294
00:17:29,440 --> 00:17:30,920
Minus F of a.
295
00:17:30,920 --> 00:17:32,810
So what's minus F of a?
296
00:17:32,810 --> 00:17:36,860
That will be plus 0.06.
297
00:17:36,860 --> 00:17:42,250
Divided by F prime of a,
the slope, which was 6.
298
00:17:42,250 --> 00:17:43,010
What do I get?
299
00:17:43,010 --> 00:17:46,680
I did 0.01.
300
00:17:46,680 --> 00:17:51,270
Actually, the two examples,
of course, are parallel.
301
00:17:51,270 --> 00:17:55,880
In a way, the graph of that
square root function kind of
302
00:17:55,880 --> 00:18:01,390
got just flipped to this graph
of the x squared function.
303
00:18:01,390 --> 00:18:03,930
The square root function and
the x squared function are
304
00:18:03,930 --> 00:18:06,980
just sort of inverses
to each other.
305
00:18:06,980 --> 00:18:14,370
Their graphs flip and so now the
slope is 6 instead of 1/6.
306
00:18:14,370 --> 00:18:22,940
And do you see that that
distance, this is the x minus
307
00:18:22,940 --> 00:18:29,320
a distance that has
to be 0.01.
308
00:18:29,320 --> 00:18:31,930
That's what I concluded.
309
00:18:31,930 --> 00:18:38,360
I go across 1 if I'm going
up/down by 6 because
310
00:18:38,360 --> 00:18:42,910
the slope is 6.
311
00:18:42,910 --> 00:18:45,250
OK, good.
312
00:18:45,250 --> 00:18:47,390
Now, how close am I?
313
00:18:47,390 --> 00:18:50,440
Well, I can't resist asking.
314
00:18:50,440 --> 00:18:58,010
Suppose I multiply
3.01 by 3.01.
315
00:18:58,010 --> 00:19:04,450
Am I close to the 9.06 that
was my whole goal?
316
00:19:04,450 --> 00:19:05,680
And how close?
317
00:19:05,680 --> 00:19:07,550
Of course, I'm not going
to be exact.
318
00:19:07,550 --> 00:19:12,530
If I do that multiplication
I get 301, 903.
319
00:19:12,530 --> 00:19:21,810
Combined I get 9 and the decimal
put it in there, 0601.
320
00:19:21,810 --> 00:19:28,940
So by that method or by that
method I ended up with 3.01 as
321
00:19:28,940 --> 00:19:34,590
much closer to the square
root of 9.06.
322
00:19:34,590 --> 00:19:40,300
You see when I square 3.01,
it slightly overshot.
323
00:19:40,300 --> 00:19:43,800
It slightly overshot the 9.06.
324
00:19:43,800 --> 00:19:50,030
This little error there, this
is the overshoot error.
325
00:19:50,030 --> 00:19:55,000
And when I square it, it's way
out in the ten thousandths
326
00:19:55,000 --> 00:19:59,890
place, the fourth
decimal place.
327
00:19:59,890 --> 00:20:00,010
OK.
328
00:20:00,010 --> 00:20:02,240
So those are the two
parallel examples.
329
00:20:02,240 --> 00:20:07,700
May I go back to the linear
approximation?
330
00:20:07,700 --> 00:20:11,370
Because I think another
example's appropriate there.
331
00:20:11,370 --> 00:20:14,450
And then I'll come back and give
another Newton example.
332
00:20:14,450 --> 00:20:17,250
So that's my plan here.
333
00:20:17,250 --> 00:20:19,620
Two examples of each.
334
00:20:19,620 --> 00:20:22,480
So those examples were parallel,
the next two
335
00:20:22,480 --> 00:20:25,520
examples will be a
little different.
336
00:20:25,520 --> 00:20:29,420
So let me show you
example two.
337
00:20:29,420 --> 00:20:42,620
So linear approximation
example two.
338
00:20:42,620 --> 00:20:49,160
339
00:20:49,160 --> 00:20:49,298
So I want--
340
00:20:49,298 --> 00:20:50,440
OK, I'm going to look
for something that
341
00:20:50,440 --> 00:20:52,770
I don't know exactly.
342
00:20:52,770 --> 00:20:58,980
Let me take e to the 0.01.
343
00:20:58,980 --> 00:21:06,160
What's the value of e to
the power of 0.01?
344
00:21:06,160 --> 00:21:11,930
So the function is e to the x.
345
00:21:11,930 --> 00:21:17,080
346
00:21:17,080 --> 00:21:24,080
And I'm looking at x is 0.01.
347
00:21:24,080 --> 00:21:25,700
That's what I want.
348
00:21:25,700 --> 00:21:29,550
I am not going to get an
exact number here.
349
00:21:29,550 --> 00:21:32,960
I'm going to follow
the tangent line.
350
00:21:32,960 --> 00:21:35,710
I'm going to get
a close number.
351
00:21:35,710 --> 00:21:38,120
So where shall I start?
352
00:21:38,120 --> 00:21:43,510
I'll start with a number that's
close to that and where
353
00:21:43,510 --> 00:21:45,530
I do you know the correct e.
354
00:21:45,530 --> 00:21:48,030
And a number close
to that is take--
355
00:21:48,030 --> 00:21:52,630
I'll choose a to be 0.
356
00:21:52,630 --> 00:21:55,110
That's close to 0.01.
357
00:21:55,110 --> 00:22:00,770
Then f of a is e
to the 0 power.
358
00:22:00,770 --> 00:22:03,790
So that's 1.
359
00:22:03,790 --> 00:22:07,260
Now for the straight line
approximation, I also need the
360
00:22:07,260 --> 00:22:09,020
correct slope.
361
00:22:09,020 --> 00:22:12,360
Correct slope at a?
362
00:22:12,360 --> 00:22:14,610
Well, I do know the
slope of this.
363
00:22:14,610 --> 00:22:18,460
f prime at 0.
364
00:22:18,460 --> 00:22:19,970
That's my a.
365
00:22:19,970 --> 00:22:21,410
f prime at 0.
366
00:22:21,410 --> 00:22:23,320
0 is my a.
367
00:22:23,320 --> 00:22:25,640
Well, I know the derivative
of e to the x.
368
00:22:25,640 --> 00:22:27,130
That's one thing I like.
369
00:22:27,130 --> 00:22:29,510
It's e to the x.
370
00:22:29,510 --> 00:22:34,500
So at x equals 0, again I get a
1 for the derivative, which
371
00:22:34,500 --> 00:22:36,410
is the same, I get e to the 0.
372
00:22:36,410 --> 00:22:38,460
I get also a 1.
373
00:22:38,460 --> 00:22:42,620
So now I know what's happening
at a equals 0.
374
00:22:42,620 --> 00:22:46,740
I want to know approximately
what's happening at the nearby
375
00:22:46,740 --> 00:22:53,360
point, 0.01.
376
00:22:53,360 --> 00:22:56,780
e to the 0.01, e to the x.
377
00:22:56,780 --> 00:22:59,940
This is the e to the x.
378
00:22:59,940 --> 00:23:02,070
That's my function.
379
00:23:02,070 --> 00:23:06,740
And I'm only going to get it
approximately, is the value at
380
00:23:06,740 --> 00:23:09,930
this known point.
381
00:23:09,930 --> 00:23:18,880
The value at the known point 0,
the exact exponential is 1.
382
00:23:18,880 --> 00:23:26,900
Plus x minus a times the slope,
the corrects slope, at
383
00:23:26,900 --> 00:23:30,640
this not quite perfect
point, 0.
384
00:23:30,640 --> 00:23:32,090
And the correct slope is 1.
385
00:23:32,090 --> 00:23:34,880
386
00:23:34,880 --> 00:23:38,660
And of course, a was 0.
387
00:23:38,660 --> 00:23:42,310
So you see I'm using all the
facts at a to get an
388
00:23:42,310 --> 00:23:44,720
approximate fact at x.
389
00:23:44,720 --> 00:23:46,340
And what have I got here?
390
00:23:46,340 --> 00:23:47,590
I've just got 1 plus x.
391
00:23:47,590 --> 00:23:51,990
392
00:23:51,990 --> 00:23:54,310
You know, in a way,
that's perfect.
393
00:23:54,310 --> 00:23:58,540
Because it shows what the linear
approximation is doing.
394
00:23:58,540 --> 00:24:00,930
You remember the series
for e to the x?
395
00:24:00,930 --> 00:24:03,320
My correct function
is e to the x.
396
00:24:03,320 --> 00:24:06,900
My approximate function
is 1 plus x.
397
00:24:06,900 --> 00:24:08,450
What's the connection?
398
00:24:08,450 --> 00:24:12,750
You remember that e to the x,
the series for e to the x
399
00:24:12,750 --> 00:24:21,290
started out 1 plus x plus 1/2
x squared, 1/6 x cubed.
400
00:24:21,290 --> 00:24:25,270
Those are the guys, those are
the higher order corrections
401
00:24:25,270 --> 00:24:30,440
that following the
line misses.
402
00:24:30,440 --> 00:24:35,510
Those are the parts where the
function, the curve e to the
403
00:24:35,510 --> 00:24:38,020
x, has left the line.
404
00:24:38,020 --> 00:24:44,340
405
00:24:44,340 --> 00:24:48,410
But if I don't go to far--So
this is 1.01.
406
00:24:48,410 --> 00:24:49,980
x is 0.01.
407
00:24:49,980 --> 00:24:51,240
That's my approximation.
408
00:24:51,240 --> 00:24:52,390
1 plus x.
409
00:24:52,390 --> 00:24:55,320
That's the thing to notice.
410
00:24:55,320 --> 00:25:00,410
That linear approximations and
you could come back to that--
411
00:25:00,410 --> 00:25:04,200
the formula for any
f and any a.
412
00:25:04,200 --> 00:25:11,030
Linear approximations are just
like those power series.
413
00:25:11,030 --> 00:25:14,940
Just like the e to the x equal
1 plus x plus 1/2 x squared
414
00:25:14,940 --> 00:25:17,110
plus so on.
415
00:25:17,110 --> 00:25:22,810
Except we cut them off after
just the constant term and the
416
00:25:22,810 --> 00:25:23,810
linear term.
417
00:25:23,810 --> 00:25:28,150
That's what this linear
approximation is about.
418
00:25:28,150 --> 00:25:31,000
And you might say, what's
the next term?
419
00:25:31,000 --> 00:25:35,980
And of course we know that the
next term is 1/2 x squared and
420
00:25:35,980 --> 00:25:39,310
you could ask, what's the
next term in this?
421
00:25:39,310 --> 00:25:43,430
In the general case, actually
let me tell you the next term.
422
00:25:43,430 --> 00:25:45,630
Next would be--
423
00:25:45,630 --> 00:25:47,630
but we're not using it.
424
00:25:47,630 --> 00:25:49,660
Would be the 1/2.
425
00:25:49,660 --> 00:25:54,200
It would be the x minus a
squared and it would be the
426
00:25:54,200 --> 00:25:55,950
second derivative at a.
427
00:25:55,950 --> 00:25:59,010
428
00:25:59,010 --> 00:26:02,670
If we kept it that's
what we would keep.
429
00:26:02,670 --> 00:26:04,980
OK, good.
430
00:26:04,980 --> 00:26:08,600
You saw the main point of linear
approximation stop at
431
00:26:08,600 --> 00:26:09,670
the linear term.
432
00:26:09,670 --> 00:26:14,690
Finally, I go back to an example
of Newton's method.
433
00:26:14,690 --> 00:26:16,900
A second example of
Newton's method.
434
00:26:16,900 --> 00:26:19,990
So I've been thinking,
what should I do?
435
00:26:19,990 --> 00:26:24,520
Let me use Newton's method
the way it's really used.
436
00:26:24,520 --> 00:26:29,840
The way you use Newton's method
is you do this to come
437
00:26:29,840 --> 00:26:32,360
close to the solution.
438
00:26:32,360 --> 00:26:35,360
And then, you do it again.
439
00:26:35,360 --> 00:26:40,140
You do it again starting
at 3.01.
440
00:26:40,140 --> 00:26:44,170
So I planned to do just the same
thing for the next step
441
00:26:44,170 --> 00:26:48,135
of Newton's method,
except the a.
442
00:26:48,135 --> 00:26:52,210
443
00:26:52,210 --> 00:26:59,810
I'm still aiming to solve this
same equation, but I'm going
444
00:26:59,810 --> 00:27:02,130
to get closer than 3.01.
445
00:27:02,130 --> 00:27:05,090
I got closer than 3 to 3.01.
446
00:27:05,090 --> 00:27:07,660
Now I'm going to
restart there.
447
00:27:07,660 --> 00:27:11,150
a is now going to be 3.01.
448
00:27:11,150 --> 00:27:16,350
I need to compute F of a
for Newton's method.
449
00:27:16,350 --> 00:27:20,900
So I have to do 3.01 squared and
take away that to see how
450
00:27:20,900 --> 00:27:22,350
wrong I am.
451
00:27:22,350 --> 00:27:22,470
HA.
452
00:27:22,470 --> 00:27:23,980
We did 3.01 squared.
453
00:27:23,980 --> 00:27:26,300
Actually, right there.
454
00:27:26,300 --> 00:27:30,970
So if I take away the 9.06, the
F of a is-- well, that was
455
00:27:30,970 --> 00:27:32,330
the whole point.
456
00:27:32,330 --> 00:27:34,910
That it was pretty darn close.
457
00:27:34,910 --> 00:27:37,750
But nothing compared to the
closeness we're going to get
458
00:27:37,750 --> 00:27:39,600
at the second term.
459
00:27:39,600 --> 00:27:40,850
And what's F prime of a?
460
00:27:40,850 --> 00:27:44,420
461
00:27:44,420 --> 00:27:46,650
Take the derivative 2x.
462
00:27:46,650 --> 00:27:49,350
At the point a it's 2a.
463
00:27:49,350 --> 00:27:52,560
And a is now 3.01.
464
00:27:52,560 --> 00:27:56,030
So the slope is 6.02.
465
00:27:56,030 --> 00:27:57,720
You see what I'm doing.
466
00:27:57,720 --> 00:28:05,310
I'm just moving over to this
point, which that has become
467
00:28:05,310 --> 00:28:10,470
now the a in the second try.
468
00:28:10,470 --> 00:28:13,230
a in the second cycle
of Newton's method.
469
00:28:13,230 --> 00:28:15,760
This is how Newton's method
really is used.
470
00:28:15,760 --> 00:28:21,170
And now let's find the new
x, the highly improved
471
00:28:21,170 --> 00:28:24,470
x, better than 3.01.
472
00:28:24,470 --> 00:28:29,190
So Newton's method says
x, the new x, minus a.
473
00:28:29,190 --> 00:28:32,160
I'm just using Newton's
formula.
474
00:28:32,160 --> 00:28:37,280
x minus the a is supposed
to be minus the F of a.
475
00:28:37,280 --> 00:28:47,340
So that's 0.0001 divided
by F prime of a, 6.02.
476
00:28:47,340 --> 00:28:50,330
This is the delta
x you could say.
477
00:28:50,330 --> 00:28:52,000
This is the little correction.
478
00:28:52,000 --> 00:28:52,810
It's negative.
479
00:28:52,810 --> 00:28:55,890
It means that we need to
pull back a little.
480
00:28:55,890 --> 00:28:57,130
And you see that.
481
00:28:57,130 --> 00:29:02,260
We slightly, slightly overshot
by following the tangent line.
482
00:29:02,260 --> 00:29:05,050
The curve went up a little
across 0 a little before the
483
00:29:05,050 --> 00:29:07,020
tangent line.
484
00:29:07,020 --> 00:29:08,770
This is extremely close.
485
00:29:08,770 --> 00:29:17,980
So now this gives me the new x
right here, 3.01 minus this
486
00:29:17,980 --> 00:29:20,360
tiny little bit.
487
00:29:20,360 --> 00:29:25,170
And so that's what the
calculator will do.
488
00:29:25,170 --> 00:29:27,120
I hope you'll do it
on a calculator.
489
00:29:27,120 --> 00:29:31,740
Just make the calculator take
that quantity, then
490
00:29:31,740 --> 00:29:33,030
make it square it.
491
00:29:33,030 --> 00:29:35,610
492
00:29:35,610 --> 00:29:38,740
Then just go through this.
493
00:29:38,740 --> 00:29:39,910
Find the new x.
494
00:29:39,910 --> 00:29:40,530
Square it.
495
00:29:40,530 --> 00:29:41,840
Subtract 9.06.
496
00:29:41,840 --> 00:29:43,560
Let's see how close it is.
497
00:29:43,560 --> 00:29:45,260
I believe that the error--
498
00:29:45,260 --> 00:29:48,450
499
00:29:48,450 --> 00:29:51,430
I don't know what it is.
500
00:29:51,430 --> 00:29:55,930
That's more than I can do in
my head, squaring that 3.01
501
00:29:55,930 --> 00:29:57,880
minus this little tiny bit.
502
00:29:57,880 --> 00:30:07,590
But I am confident that the
error, the x new squared.
503
00:30:07,590 --> 00:30:13,010
This is the formula for
x new, the second
504
00:30:13,010 --> 00:30:15,570
cycle of Newton's method.
505
00:30:15,570 --> 00:30:21,340
I think that minus the
9.06, I don't know--
506
00:30:21,340 --> 00:30:22,915
can I just put a
bunch of zeros?
507
00:30:22,915 --> 00:30:28,010
508
00:30:28,010 --> 00:30:30,270
Somebody will want me to
put in a 1 here, so
509
00:30:30,270 --> 00:30:32,410
I'll put in a 1.
510
00:30:32,410 --> 00:30:35,320
I bet it's way out there.
511
00:30:35,320 --> 00:30:38,480
So Newton's method is really
a terrific success.
512
00:30:38,480 --> 00:30:41,370
Follow the line, then follow
the next tangent line.
513
00:30:41,370 --> 00:30:46,260
Then follow the next one and you
home in very, very quickly
514
00:30:46,260 --> 00:30:48,560
on the exact answer.
515
00:30:48,560 --> 00:30:52,400
You get more and more decimal
places correct.
516
00:30:52,400 --> 00:30:55,630
OK, that's two uses
of calculus coming
517
00:30:55,630 --> 00:30:58,080
from the same idea.
518
00:30:58,080 --> 00:31:02,120
The same idea delta
f over delta x.
519
00:31:02,120 --> 00:31:06,070
In one case, it was f
that we didn't know.
520
00:31:06,070 --> 00:31:09,440
In this case, it was x
that we didn't know.
521
00:31:09,440 --> 00:31:13,390
In both cases that formula
gives a terrific and a
522
00:31:13,390 --> 00:31:18,010
terrifically simple and
close approximation
523
00:31:18,010 --> 00:31:20,030
to the exact answer.
524
00:31:20,030 --> 00:31:21,210
Good.
525
00:31:21,210 --> 00:31:23,290
Thank you very much.
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00:31:23,290 --> 00:31:25,100
ANNOUNCER: This has been
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527
00:31:25,100 --> 00:31:27,490
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00:31:27,490 --> 00:31:29,760
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00:31:29,760 --> 00:31:30,980
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00:31:30,980 --> 00:31:34,110
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